JW100G1 Lineage Power, JW100G1 Datasheet - Page 13

• Manufacturer Part Number: JW100G1
• Manufacturer: Lineage Power

Data Sheet
March 2008
Thermal Considerations
Heat Transfer with Heat Sinks
Thermal derating with heat sinks is expressed by using
the overall thermal resistance of the module. Total
module thermal resistance (θca) is defined as the max-
imum case temperature rise (ΔT
module power dissipation (P
The location to measure case temperature (T
shown in Figure 22. Case-to-ambient thermal resis-
tance vs. airflow is shown, for various heat sink config-
urations and heights, in Figure 26. These curves were
obtained by experimental testing of heat sinks, which
are offered in the product catalog.
Figure 26. Case-to-Ambient Thermal Resistance
Lineage Power
θ
ca
8
7
6
5
4
3
2
1
0
0
0
=
Curves; Either Orientation
ΔT
-------------------- -
AIR VELOCITY MEASURED IN m/s (ft./min.)
(100)
0.5
P
C max
,
D
(200)
1.0
=
(
----------------------- -
T
D
(300)
C
):
P
1.5
–
D
C, max
T
1 1/2 IN HEAT SINK
1 IN HEAT SINK
1/2 IN HEAT SINK
1/4 IN HEAT SINK
NO HEAT SINK
A
(continued)
)
(400)
) divided by the
2.0
(continued)
JW100G and JW150G Power Modules; dc-dc Converters:
(500)
2.5
C
36 Vdc to 75 Vdc Input, 2.5 Vdc Output; 50 W to 75 W
) is
8-1153 (F)
(600)
3.0
These measured resistances are from heat transfer
from the sides and bottom of the module as well as the
top side with the attached heat sink; therefore, the
case-to-ambient thermal resistances shown are gener-
ally lower than the resistance of the heat sink by itself.
The module used to collect the data in Figure 26 had a
thermal-conductive dry pad between the case and the
heat sink to minimize contact resistance. The use of
Figure 26 is shown in the following example.
Example
If an 85 °C case temperature is desired, what is the
minimum airflow necessary? Assume the JW150G
module is operating at V
of 30 A, maximum ambient air temperature of 40 °C,
and the heat sink is 1/2 inch.
Solution
Given: V
Determine P
Then solve the following equation:
Use Figure 26 to determine air velocity for the 1/2 inch
heat sink.
The minimum airflow necessary for the JW150G
module is 2.1 m/s (420 ft./min.).
θ
θ
θ
ca
ca
ca
I
T
T
Heat sink = 1/2 inch
P
O
=
=
=
A
C
I
D
= 55 V
= 30 A
= 40 °C
= 85 °C
= 23.5 W
1.9 °C/W
D
(
----------------------- -
(
----------------------- -
T
85 40
by using Figure 25:
23.5
C
P
–
–
D
T
A
)
)
I
= 55 V and an output current
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